Question

Evaluate \(f_{x}, f_{y}\), and \(f_{z}\) at the given point. \(f(x, y, z)=x^{2} y^{3}+2 x y z-3 y z, \quad(-2,1,2)\)

Short answer

The evaluated partial derivatives \(f_x, f_y,\) and \(f_z\) at the point (-2,1,2) are -4, 8, and -7 respectively.

Step-by-step solution

Differentiate f with respect to x

Here, y and z are treated as constant. Differentiating \(x^{2} y^{3}\) gives \(2x y^{3}\) (using power rule), and differentiating \(2x y z\) gives \(2 y z\) (since derivative of x is 1), while differentiating \(–3y z\) gives \(0\) because that term does not contain x.

Differentiate f with respect to y

Now, treat x and z as constants. Differentiating \(x^{2} y^{3}\) gives \(3x^{2} y^{2}\), and differentiating \(2 x y z\) gives \(2 x z\), while differentiating \(–3y z\) gives \(-3 z\).

Differentiate f with respect to z

Finally, treat x and y as constants. Only the terms \(2 x y z\) and \(–3y z\) contain z. Differentiating those two terms gives \(2 x y\) and \(-3 y\), respectively.

Evaluate f_x, f_y and f_z at the point

Substitute the coordinates (-2, 1, 2) into each of the three derivatives found. The derivative \(f_x\) at point (-2, 1, 2) is \(2(-2)(1)^3 = -4\). The derivative \(f_y\) at point (-2, 1, 2) is \((3)(-2)^2(1)^2 + 2(-2)(2) - 3(2) = 8\). The derivative \(f_z\) at point (-2, 1, 2) is \(2(-2)(1) - 3(1) = -7\).

Key concepts

Multivariable Calculus

Multivariable calculus is an expansion of calculus to functions with more than one variable. Unlike single-variable calculus, where there's just one independent variable, multivariable calculus deals with functions involving several variables such as f(x, y, z). This type of calculus is crucial in fields like physics, engineering, economics, and many more where multiple factors are at play simultaneously.

When working with functions of multiple variables, we often need to understand how the function changes in response to changes in just one of the variables, keeping the others fixed. This is where the concept of partial differentiation comes into play. In the step-by-step solution provided, partial derivatives are calculated with respect to each variable: x, y, and z. This process involves treating all other variables as constants and differentiating with respect to the variable of interest, effectively investigating the slope of the function in the direction of that variable.

Power Rule Differentiation

The power rule is among the most fundamental tools in calculus, making the process of differentiation swift and straightforward. Applicable to both single-variate and multivariate functions, the power rule states that the derivative of x^n with respect to x is nx^(n-1). Here, n is any real number, representing the power to which the variable is raised.

In the context of partial differentiation for multivariable calculus, when applying the power rule, any variable other than the one with respect to which you're differentiating should be considered a constant. For example, differentiating f(x, y, z) = x^2 y^3 with respect to x uses the power rule and treats y^3 as a constant, resulting in 2xy^3, which signifies the rate at which f changes with respect to x while holding y and z constant.

Chain Rule Differentiation

The chain rule in calculus is a formula to compute the derivative of a composite function. In other words, when one function is nested within another, the chain rule allows us to differentiate such functions systematically. The chain rule is expressed in the form (f(g(x)))' = f'(g(x)) * g'(x), where f and g are functions of x.

When it comes to multivariable calculus, the chain rule becomes even more powerful. It’s used when a function's variables themselves are functions of other variables or when applying partial differentiation in complex scenarios. For instance, if you have a function z = f(x, y) where x and y are functions of another variable t, then the derivative of z with respect to t would involve partial derivatives of f with respect to x and y, and derivatives of x and y with respect to t. This extended form helps to unravel how changes in t affect z, via the paths of x and y.